Q.In a permanent magnet at room temperature

Concept & Intuition

In amplitude modulation, the carrier wave’s amplitude varies between a maximum $A_{\text{max}}$ and a minimum $A_{\text{min}}$ as the message signal changes. The modulation index $m$ measures how much the carrier is “swung” by the message. It is defined as

The phrase “400% more than” is the key trap: it means the maximum is the minimum plus 400% of the minimum, not 400% of the minimum. That is, $A_{\text{max}} = A_{\text{min}} + 4A_{\text{min}} = 5A_{\text{min}}$.

Step-by-step solution

1. Interpret the given condition “Maximum amplitude is 400% more than its minimum amplitude” means:

So the maximum is five times the minimum.

2. Write the modulation index formula For AM, the modulation index (or depth of modulation) is:

3. Substitute the relation Replace $A_{\text{max}}$ with $5A_{\text{min}}$:

4. Simplify Cancel $A_{\text{min}}$ (it is nonzero):

Concept & Intuition

A transformer works by changing magnetic flux in its iron core. That changing flux doesn’t just induce a voltage in the secondary coil — it also induces voltages inside the core itself. Because the core is a conductor, these voltages drive circulating currents called eddy currents. They flow in closed loops, dissipating power as heat (like a tiny electric heater). The goal is to limit these currents without ruining the core’s magnetic properties. The trick: make the core from thin, insulated sheets (laminations) instead of one solid block. Each sheet is so thin that the induced voltage across it is tiny, and the insulation between sheets blocks large cross‑section loops.

Step‑by‑step reasoning

1. Why eddy currents form

   Faraday’s law says a changing magnetic field $\frac{d\Phi}{dt}$ induces an emf in any closed loop. In a solid iron core, the core itself is a giant set of nested conducting loops. The induced emf drives currents whose magnitude depends on the loop’s resistance $R$. Power loss is $P = I^2 R$, so we want to either reduce the induced emf or increase the resistance.

2. How laminations help

   Laminating means stacking thin sheets of iron, each coated with an insulating varnish. The sheets are oriented parallel to the magnetic field. Now the induced emf appears across each thin sheet, but the loop area for a current is drastically smaller. Since induced emf ∝ area of the loop, the voltage driving eddies in each sheet is much smaller. Also, the insulation forces currents to stay within a single sheet, so the effective resistance is higher (the current path is long and narrow). Both effects reduce eddy current loss dramatically — typically to about $1/n^2$ of the solid‑core loss for $n$ laminations.

3. Why the other options fail

   • (A) Winding primary and secondary one over the other improves magnetic coupling (reduces leakage flux) but does nothing to stop eddy currents in the core.
   • (B) Using thick wire reduces resistive losses in the coils (copper losses), but eddy currents are in the core, not the wire.
   • (D) Low hysteresis loss material reduces a different core loss (hysteresis — energy lost magnetising and demagnetising the domains), but eddy current loss is a separate phenomenon. You can have low hysteresis and still huge eddy currents if the core is solid.

4. Classic pitfall

   Watch out

   Students often confuse “eddy current loss” with “copper loss” or “hysteresis loss.” Remember: eddy currents are induced in the core; laminations break them up. Hysteresis loss is reduced by using silicon steel or other soft magnetic materials, not by laminating.

5. Neat insight

   Tip

   Laminations work because the induced voltage scales with the square of the lamination thickness (for a given frequency and flux density). Halving the thickness quarters the eddy current loss. That’s why transformer cores are made of many thin sheets rather than a few thick ones.

The key idea here is that eddy currents are loops of current induced within a solid conductor when it is placed in a changing magnetic field. In a transformer, the iron core experiences a rapidly alternating magnetic flux. This flux induces a voltage in the core material itself, just as it does in the secondary coil. Because the core is a conductor, this voltage drives large circulating currents — eddy currents — which dissipate energy as heat (I²R loss) and waste power.

The question asks which method directly reduces these eddy current losses. Let’s examine each option.

1. Option (A): Winding the primary and secondary coils one over the other. This is a common practice to improve magnetic coupling (leakage flux is reduced), which increases efficiency. But it does nothing to stop currents from flowing inside the core. The eddy currents are in the core, not in the windings, so this doesn’t address the problem.

2. Option (B): Using thick wire. Thicker wire reduces the resistance of the coils, which cuts down on copper losses (I²R heating in the windings). Eddy current losses occur in the core, not the coils. The wire gauge has no effect on the core’s internal currents.

3. Option (C): Using magnetic material with low hysteresis loss. Hysteresis loss is a different beast — it’s the energy spent to repeatedly magnetise and demagnetise the core material as the field reverses. A material with a narrow hysteresis loop (like silicon steel) reduces this loss. But hysteresis and eddy currents are separate mechanisms. A low-hysteresis material doesn’t automatically suppress eddy currents; you need a different strategy for that.

4. Option (D): Using a laminated core. This is the classic solution. The core is built from thin sheets (laminations) stacked together, each coated with a thin insulating layer (like varnish). The insulation breaks the conducting path across the thickness of the core. Eddy currents try to flow in large loops, but the laminations confine them to tiny loops within each thin sheet. Since the area of each loop is drastically reduced, the induced emf (which is proportional to area) drops, and so the eddy current magnitude plummets. The power loss, proportional to the square of the current, becomes negligible.

Concept & Intuition

When two long straight parallel wires carry current in the same direction, the magnetic fields they produce at a point between them point in opposite directions. Why? Use the right-hand rule: for a wire, the field circles around it. Between the wires, the field from wire A goes one way (say, into the page) and from wire B the opposite way (out of the page). So the net field is the difference of their magnitudes. The field from a long straight wire is given by $B = \frac{\mu_0 I}{2\pi r}$, where $r$ is the perpendicular distance from the wire.

Step-by-step solution

1. Identify distances

   Wires A and B are 5 m apart. The point is 2 m from wire A, between the wires. Therefore, its distance from wire B is $5 - 2 = 3$ m.

2. Compute field due to wire A

Direction: using right-hand rule (thumb along current, fingers curl), at a point to the right of wire A (since B is to the right), the field points into the page (say, negative direction).

3. Compute field due to wire B

Direction: at a point to the left of wire B (since A is to the left), the field points out of the page (positive direction).

4. Net field Since the fields are opposite, subtract the smaller from the larger:

The net direction is out of the page (the direction of the stronger field from wire B).

Concept of Mutual Inductance and Flux Linkage

When two coils are placed close to each other, a changing current in one coil can induce an electromotive force (EMF) in the other coil. This phenomenon is called mutual induction. The strength of this coupling between the coils is quantified by a property called mutual inductance, denoted by $M$.

The magnetic field produced by the current in the first coil passes through the second coil, creating a magnetic flux linkage with it. The key idea is that the magnetic flux linkage ($\Phi_2$) with the second coil is directly proportional to the current ($I_1$) flowing in the first coil.

If the current in the first coil changes, the magnetic flux linkage with the second coil also changes proportionally. Therefore, the change in flux linkage ($\Delta \Phi_2$) is related to the change in current ($\Delta I_1$) by:

$\Delta \Phi_2 = M \Delta I_1$

Step-by-Step Solution

1. Identify the given values:

   We are given the mutual inductance ($M$) of the pair of coils and the initial and final currents in one of the coils.

   • Mutual inductance, $M = 2\,\mathrm{H}$
   • Initial current, $I_{\text{initial}} = 0\,\mathrm{A}$
   • Final current, $I_{\text{final}} = 25\,\mathrm{A}$
   • Time taken for the change, $\Delta t = 1\,\mathrm{s}$

2. Determine the change in current:

   The current in one coil changes from $0\,\mathrm{A}$ to $25\,\mathrm{A}$.

   The change in current, $\Delta I$, is the final current minus the initial current:

   $\Delta I = I_{\text{final}} - I_{\text{initial}} = 25\,\mathrm{A} - 0\,\mathrm{A} = 25\,\mathrm{A}$

3. Apply the formula for change in flux linkage:

   Using the relationship $\Delta \Phi = M \Delta I$, we can calculate the change in flux linkage with the other coil.

   $\Delta \Phi = (2\,\mathrm{H}) \times (25\,\mathrm{A})$

   $\Delta \Phi = 50\,\mathrm{Wb}$

   Watch out

   The time duration ($\Delta t = 1\,\mathrm{s}$) given in the problem is relevant if we were asked to calculate the induced EMF ($E = -M \frac{\Delta I}{\Delta t}$). However, for calculating the change in flux linkage ($\Delta \Phi$), the time taken for the current change is not required. It is a common distractor.

The change of flux linkage with the other coil is $50\,\mathrm{Wb}$.

The key insight is that the electron moves in a circle perpendicular to the solenoid axis. Inside a long solenoid, the magnetic field is uniform and parallel to the axis. So the electron’s velocity is perpendicular to the field — a perfect setup for circular motion under the Lorentz force.

The magnetic force $q v B$ acts as the centripetal force $m v^2 / r$. This directly gives us $B$, the field inside the solenoid. Then we use the solenoid formula $B = \mu_0 n I$, where $n$ is the number of turns per unit length, to find $I$.

1. Relate the magnetic force to circular motion. For an electron moving perpendicular to a uniform magnetic field,

Cancel one factor of $v$ (the electron is moving, so $v \neq 0$):

2. Plug in the numbers. $m = 9 \times 10^{-31}$ kg, $v = 4.4 \times 10^6$ m/s, $q = 1.6 \times 10^{-19}$ C, $r = 2.5$ cm $= 0.025$ m.

First compute numerator: $9 \times 4.4 = 39.6$, so $39.6 \times 10^{-25}$.

Denominator: $1.6 \times 0.025 = 0.04$, so $0.04 \times 10^{-19} = 4 \times 10^{-21}$.

3. Now use the solenoid formula. The solenoid has 70 turns/cm. Convert to SI: $70$ turns/cm $= 7000$ turns/m. So $n = 7000$ m$^{-1}$.

4. Simplify. $4\pi \times 10^{-7} \times 7000 = 4\pi \times 7 \times 10^{-4} = 28\pi \times 10^{-4}$. So

Using $\pi \approx 3.14$, $28\pi \approx 87.92$. Then

The exact calculation: $\frac{9.9}{28\pi} = \frac{9.9}{87.9646} \approx 0.1125$ A, which is exactly 112.5 mA.

The core idea is superposition of magnetic fields. A long solenoid gives a uniform axial field inside, while a straight current along its axis gives a circumferential (azimuthal) field that varies with radial distance. At a point 5 cm from the axis, these two fields are perpendicular — one along the solenoid’s axis, the other tangent to a circle around the wire. The resultant field’s direction is the vector sum, so the angle it makes with the axis is determined by the ratio of the two field magnitudes.

Let’s work it through.

1. Solenoid field The solenoid has $n = 10.0$ turns/cm $= 1000$ turns/m. Current $I_s = 7$ mA $= 7 \times 10^{-3}$ A. The axial field inside a long solenoid is

Numerically, $28 \times 3.14 \times 10^{-7} \approx 8.79 \times 10^{-6}$ T. Keep it symbolic for now.

2. Wire field The straight conductor along the axis carries current $I_w$ (unknown). At a radial distance $r = 5$ cm $= 0.05$ m, the circumferential field is

(since $2/0.05 = 40$, so $40 \times 10^{-7} = 4 \times 10^{-6}$).

3. Resultant direction The two fields are perpendicular: $B_s$ along the axis, $B_w$ tangential. The resultant makes angle $\theta$ with the axis, where

Given $\theta = 60^\circ$, $\tan 60^\circ = \sqrt{3} \approx 1.7$. So

4. Solve for $I_w$ Substitute the expressions:

Simplify the denominator: $28\pi \times 10^{-7} = 2.8\pi \times 10^{-6}$. So

Using $\pi \approx 3.14$, $\sqrt{3} \approx 1.7$:

The magnetic field inside a long solenoid is a fundamental concept in electromagnetism. Understanding its dependence on various parameters is key to solving this problem.

Concept and Intuition

A solenoid is essentially a coil of wire wound in a tightly packed helix. When current flows through this wire, it creates a magnetic field. For a long solenoid (where its length is much greater than its radius), the magnetic field inside it is remarkably uniform and directed along its axis.

The strength of this magnetic field at the center of a long solenoid depends on two main factors:

1. The current ($I$) flowing through the wire: A larger current produces a stronger magnetic field.
2. The number of turns per unit length ($n$) of the solenoid: More turns packed into a given length means a denser winding, leading to a stronger magnetic field.

Crucially, for a long solenoid, the magnetic field at its center does not depend on the total length of the solenoid or the total number of turns individually, but rather on their ratio, which is the turns per unit length.

When a solenoid is cut into two equal parts, both its total length and its total number of turns are halved. However, the ratio of turns to length, which is $n$, remains constant. Since the current also remains the same, the magnetic field at the center of each new solenoid will be identical to the original.

Step-by-step derivation:

1. Initial Solenoid: Let the original solenoid have a total length $L$ and a total number of turns $N$. The number of turns per unit length for the original solenoid is $n = \frac{N}{L}$. Let the current flowing through the solenoid be $I$. The magnetic field at the center of this original solenoid is given by:

2. Cutting the Solenoid: The solenoid is cut into two equal parts. Consider one of these new solenoids. Its new length, $L'$, will be half of the original length:

Since the solenoid is cut into two equal parts, the total number of turns in one of these new solenoids, $N'$, will also be half of the original total turns:

The current flowing through the new solenoid remains the same:

3. Magnetic Field in the New Solenoid: Now, let's calculate the number of turns per unit length for one of the new solenoids, $n'$:

Simplifying this expression:

Notice that $n'$ is equal to the original number of turns per unit length, $n$.
Now, we can find the magnetic field $B'$ at the center of the new solenoid using the same formula:

Substitute $n' = n$ and $I' = I$:

4. Comparison: Comparing the magnetic field of the new solenoid ($B'$) with the original magnetic field ($B$):

Therefore, $B' = B$.

Why this approach works

A bar magnet is a magnetic dipole. When the observation point lies far from the magnet along its axis (the "short magnet" or "point dipole" limit), the field depends only on the magnetic moment $M$ and the distance $r$. The axial field formula encodes this inverse-cube dependence, and knowing $B$ at a given $r$ lets us solve backward for $M$.

The key is recognizing that "short bar magnet" signals the use of the dipole approximation, where the magnet's physical length is negligible compared to the observation distance.

Step-by-step solution

1. Write the axial field formula for a short bar magnet. On the axis of a magnetic dipole, at distance $r$ from the center, the magnetic field is

where $M$ is the magnetic moment in $\mathrm{Am}^2$ and $\mu_0 = 4\pi \times 10^{-7}\,\mathrm{T\,m/A}$.

2. Substitute the known values. We are given $B = 10^{-10}\,\mathrm{T}$ and $r = 10\,\mathrm{m}$. Plug these in:

The $4\pi$ cancels immediately:

3. Simplify and solve for $M$. Rewrite $1000 = 10^3$:

Divide both sides by $2 \times 10^{-10}$: